Запишем область допустимых значений: \(x-1 \ne 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \ne 1.\)
Пусть \(\dfrac{1}{{x-1}} = t,\) тогда уравнение примет вид:
\({t^2} + 2t-3 = 0;\,\,\,\,\,\,D = {2^2}-4 \cdot 1 \cdot \left( {-3} \right) = 16;\,\,\,\,\,\,\,\left[ \begin{array}{l}t = \dfrac{{-2 + 4}}{2} = 1,\\t = \dfrac{{-2-4}}{2} = -3.\end{array} \right.\)
Возвращаясь к прежней переменной, получим:
\(\left[ \begin{array}{l}\dfrac{1}{{x-1}} = 1,\\\dfrac{1}{{x-1}} = -3\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x-1 = 1,\\x-1 = -\dfrac{1}{3}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = 2,\\x = \dfrac{2}{3}.\end{array} \right.\)
Ответ: \(2;\,\,\,\,\dfrac{2}{3}.\)